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A human male and female couple with normal colored ears discover that, in both of their families, their fathers (who have normal ears) each had siblings with red ears. Red ears is a rare autosomal recessive disorder. None of their grandparents had red ears. Assume that the red ears in both families were caused by mutations in the same gene, that there were no new mutations within the pedigrees, and that the mothers of the couple were not carriers of the disease gene.

A) based on the information, what is the probability that the male in the couple is a carrier of the red ear disease gene?

B) if the couple had 1 child, what would be the probability that child would have red ears?

I was discussing this problem over with a friend and we got different answers.

A)I think the answer is 25%.  I multiplied 1/2 and 1/2.
B)6.25%.  I multiplied 1/4 and 1/4.
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Is red turning to green a copy error or is it another distinct phenotype? –  Rory M Mar 11 '12 at 23:11
    
@RoryM i mixed up the colors. the question is edited –  Kirby Mar 11 '12 at 23:29
    
What's the question? The probability that this male and female will have a child with red ears? –  leonardo Mar 11 '12 at 23:45
    
@leonardo updated question –  Kirby Mar 11 '12 at 23:46
    
Great! Now we can help. :) –  leonardo Mar 12 '12 at 0:02

1 Answer 1

up vote 6 down vote accepted

Let's take Question A:

Both fathers have siblings with red ears, and red ears are an autosomal recessive trait. The grandparents did not have red ears, we know then that they were carriers of the recessive allele. Each grandparent was Nr (for Normal and red alleles respectively).

The fathers have normal ears, so they could be NN (probability 0.33) or Nr (probability 0.67).

The answer then is that each father independently has a 2/3 chance (0.67 probability) of carrying the red ears gene.

The male's mother, being RR, makes the male a carrier only if the male's father is himself a carrier. To put it mathematically, he has a 1/2 chance of being a carrier with 2/3 probability that he was a carrier to begin with. (2/3)*(1/2) = 1/3.

Therefore the male can be carrier with 1/3 chance.

Now Question B:

Short answer:

Since the male and female have symmetrical pedigrees, and we just solved the chance of the male being a carrier with 1/3 chance, then the chance of them having a child with red ears is (1/3)*(1/3) = 1/9.

Long answer:

The male's and female's fathers may each be heterozygous or homozygous normal, given the information, with both mothers being homozygous normal (NN).

If Father A is NN (crossed with Mother A, also NN), there is a 0 probability of passing on the red ears allele.

If Father A is Nr, then Person A has a 0.5 probability of carrying the red ear allele.

Person A can now be NN (with 0.25 probability) or Nr (with 0.5 probability).

Likewise with Person B.

Now lets consider all 4 combinations of genotypes that Person A and B can have when they have a child. I will write a table with one Person on each side of the table, with the probability of their genotype written in brackets beside. Each intersection will represent the frequency of carrying the red ears allele.

                Person B
             (1/3)     (2/3)
Person A      NN        Nr
(1/3) NN      0/4       2/4
(2/3) Nr      2/4       2/4

Carrying the red ear gene: (A more interesting example) There are a total of 6 outcomes in which the child carries the red ears allele, but these must be weighted by the probability that each parent has the associated genotype.

Probability of carrying 'r' = [ (1/3)(1/3)(0/4) + (1/3)(2/3)(2/4) + (2/3)(1/3)(2/4) + (2/3)(2/3)(2/4) ] / [ (1/3)(1/3) + (1/3)(2/3) + (2/3)(1/3) + (2/3)(2/3) ] = 0.25

Having a child with red ears: Probability of having red ears = [ (1/3)(1/3)(0/4) + (1/3)(2/3)(0/4) + (2/3)(1/3)(0/4) + (2/3)(2/3)(1/4) ] / [ (1/3)(1/3) + (1/3)(2/3) + (2/3)(1/3) + (2/3)(2/3) ] = 1/9 or about 0.11

It's left as an exercise to the student to derive the remaining probabilites. Nr 4/9 NN 4/9 rr 1/9

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question A asks for the probability that the male in the couple is a carrier. the male's grandparents must have been Rr (R=red ear dominant, r=recessive red ears). Therefore the male's father has a 1/2 chance of being a carrier. since the male's mother can't be a carrier, her genotype is RR. Therefore the male's parents are Rr and RR. so 1/2 * 1/2. why isn't this right? –  Kirby Mar 12 '12 at 0:43
    
You are correct that each grandparent is Rr. The male's father then can either be RR (1/4 chance) or Rr (2/4 chance) but not rr (0/4 chance). We only know the male's father has a normal phenotype. We normalize this to the possible genotypes, and then the male's father has 1/3 chance of being RR and 2/3 chance of being Rr. The male's mother, being RR, makes the male a carrier only if the male's father is himself a carrier. To put it mathematically, he has a 1/2 chance of being a carrier with 2/3 probability that he was a carrier to begin with. (2/3)*(1/2) = 1/3. –  leonardo Mar 12 '12 at 1:03
    
Made the question more clear, and corrected the solution to question A since misread it. –  leonardo Mar 12 '12 at 1:09
1  
thank you. my error was not normalizing the probabilities as rr is impossible if the person is a carrier. –  Kirby Mar 12 '12 at 1:14
    
Please be so kind as to upvote my answer and select it as the accepted answer if I've satisfied your question. Thanks. :) –  leonardo Mar 12 '12 at 1:40

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